Prooving the Tetrahedral Number Sequence?

Question

Basically, I'm doing work on Pascal's Triangle and I came upon these sequences:

The Co-efficients with x^2 went: 1 3 6 10 etc.

Through working out the sequence's differences and using the geometric nth term formula a+b(n-1) + 0.5(n-1)(n-2)

I do realise that these numbers are the triangular numbers.

This is the next sequence:

The co-efficients that have x^3: 0 1 4 10 20 etc.

I found out that the first differences of this sequence were all the triangular numbers from before. I also realise that these are the tetrahedral numbers. As well as this I discovered that the first differences of any co-efficients of x^n sequence from Pascal's Triangle form the main sequence of x^n-1, if you know what I mean.

From some research, I realised that the formula for the tetrahedral numbers is n(n+1)(n+2)/6. Or in the interest of the binomial theorem, this 6 is turned into 3!.

Anyways, I wanted to know if there is any way of 'getting to' the tetrahedral formula algebraically or with the 1st differences=main sequence of x^n-1 principal and also if this method would work when I'm trying to find out the formula for the sequence of the co-efficients of say, x^19 or something.

Thanks!

Answer 1

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Riemann Hypothesis

zeta(s) = 1/1s + 1/2s + 1/3s + ... (s = a + it) all 0's of zeta(s) in strip 0<=a<=1 lie on central line a=1/2

Twin Primes occur infinitely

Twin primes are primes that are 2 integers apart. Examples include 5 & 7, 17 & 19, 101 & 103

Goldbach's Postulate

Every even # > 2 can be expressed as the sum of 2 primes.
4=2+2, 6=3+3, 8=3+5, 10=5+5, 12=5+7, .. , 100=3+97, ...

Euclid's Parallel Postulate

Through a point, not on a line, there exists exactly 1 line parallel to the given line. (Then there's those non-Euclidean people...)